$\begin{align}  & \text{Area of rectangle }PSQR=7\times 10c{{m}^{2}}=70c{{m}^{2}} \\ & \text{Area of semicircle}=\frac{1}{2}{{\left( \frac{22}{7}\times \left( \frac{7}{2} \right) \right)}^{2}}=\frac{77}{4}c{{m}^{2}} \\ & \text{Area of the shape}=70c{{m}^{2}}+\frac{77}{4}c{{m}^{2}}=\frac{280+77}{4}c{{m}^{2}} \\ & =\frac{357}{4}c{{m}^{2}}=89.25c{{m}^{2}} \\ & \text{To the nearest whole number, the area is }89c{{m}^{2}} \\\end{align}$